(I)第n行有n个数,设a27在第n行,则 n(n−1) 2+1≤27≤ n(n+1) 2, 则n=7,且a27为第7行的第6项, 则a27=27+26-1=128+32=160; (II)依题意,第n行的总和 i i=1bij=(2n+20)+(2n+21)+…+(2n+2n-1)=(n+1)2n-1,
n i=1( i i=1bij)=b1+b2+…+bn-1+bn=(2×21-1)+(3×22-1)+…+(n×2n-1-1)+[(n+1)2n-1]=[2×21+3×22+…+n×2n-1+(n+1)2n]-n, 2 n i=1( i i=1bij)=[2×22+3×23+…+n×2n+(n+1)2n+1]-2n, 两式相减,并由等比数列前n项和公式得:
n i=1( i i=1bij)=-2×21-[22+23+…+2n]+(n+1)2n+1-n =n×(2n+1-1). 故答案为:160,n(2n+1-1)